So the surface $S$ is defined as:
$$(x^2+y^2+z^2)^2=2z\sqrt{x^2+y^2}$$
And i have to look at the object it surrounds in $\mathbb{R^3}$ So it seems I'd be using spherical coordinates in the calculation(since the left side is somehow just the equation of sphere squared). But i wonder how does the surface look like, I have to draw a sketch of it. Is it still similar to a sphere?
So using spherical coordinates gives me: $$x=r\cos\phi \cos\theta$$ $$x=r\sin\phi \cos\theta$$ $$x=r\sin\theta$$
So because of left sides sphere-like appearence for $$\phi\in[0,2\pi]$$
Also : $$r\leq \sin\theta \cos\theta(\sin\phi+\cos\phi)$$
So i need help getting bounds for $\theta$ as well as for any isnigth how to draw this object/surface. On the other hand is my approach correct? All that is left after that is calculating the integrals needed for getting the centre of mass, that i have covered.
Thank you for any help and hint.
Given the solid: $$ \Omega := \left\{ (x,\,y,\,z) \in \mathbb{R}^3 : \left(x^2+y^2+z^2\right)^2 \le 2\,z\,\sqrt{x^2+y^2}\,\right\} $$
applying the following coordinate transformation: $$ \Phi : \begin{cases} x = \rho\,\sin\varphi\,\cos\theta \\ y = \rho\,\sin\varphi\,\sin\theta \\ z = \rho\,\cos\varphi \end{cases} \; \; \; \; \; \; \text{with} \; (\rho,\,\varphi,\,\theta) \in [0,\,+\infty) \times [0,\,\pi] \times [0,\,2\pi) $$ and $J_{\Phi} = \rho^2\,\sin\varphi$, it follows that: $$ \begin{cases} \rho^4 \le \rho^2\,\sin(2\varphi) \\ \rho \ge 0 \\ 0 \le \varphi \le \pi \\ 0 \le \theta < 2\pi \end{cases} \; \; \; \; \; \; \Leftrightarrow \; \; \; \; \; \; \begin{cases} 0 \le \rho \le \sqrt{\sin(2\varphi)} \\ 0 \le \varphi \le \frac{\pi}{2} \\ 0 \le \theta < 2\pi \end{cases} $$ therefore, we have: $$ \Omega^* := \left\{ (\rho,\,\varphi,\,\theta) \in \mathbb{R}^3 : 0 \le \rho \le \sqrt{\sin(2\varphi)}, \; 0 \le \varphi \le \frac{\pi}{2}, \; 0 \le \theta < 2\pi\right\}. $$ In light of all this, we have: $$ ||\Omega|| := \iiint\limits_{\Omega} 1\,\text{d}x\,\text{d}y\,\text{d}z = \iiint\limits_{\Omega^*} J_{\Phi}\,\text{d}\rho\,\text{d}\varphi\,\text{d}\theta = \frac{\pi^2}{8} $$ $$ \overline{z} := \frac{1}{||\Omega||}\iiint\limits_{\Omega} z\,\text{d}x\,\text{d}y\,\text{d}z = \frac{1}{||\Omega||}\iiint\limits_{\Omega^*} \left(\rho\,\cos\varphi\right)J_{\Phi}\,\text{d}\rho\,\text{d}\varphi\,\text{d}\theta = \frac{4}{3\pi} $$ which is what is desired. Of course, $\overline{x} = \overline{y} = 0$ for symmetric questions, it is not necessary to integrate.